Question

Solve:-
$\underset{0}{\overset{\frac{\pi }{2}}{\int }}\log \sin xdx$

Answer 1

$I={\int }_0^{\pi /2}\log \sin xdx$

$\Rightarrow I={\int }_0^{\pi /2}\log \cos xdx$

$\Rightarrow 2I={\int }_0^{\pi /2}\log \left(\sin x\cos x\right)dx$

$={\int }_0^{\pi /2}\log \left(\frac{2\sin x\cos x}{2}\right)dx={\int }_0^{\pi /2}\log \left(\frac{\sin 2x}{2}\right)dx$

$={\int }_0^{\pi /2}\log \sin 2xdx-{\int }_0^{\pi /2}\log 2dx-----I_1$

$I_1\to$ let $2x=t\Rightarrow dx=\frac{dt}{2}$ , $x=0x=\pi /2$

$t=0t=\pi$

$I_1={\int }_0^{\pi /2}\log \sin t.\frac{dt}{2}=\frac{1}{2}{\int }_0^{\pi /2}\log \sin tdt$

As we know $\sin (\pi -x)\sin x$

$I_1=\frac{1}{2}.2{\int }_0^{\pi /2}\log \sin tdt={\int }_0^{\pi /2}\log \sin xdx$

$2I={\int }_0^{\pi /2}\log \sin xdx-\frac{\pi }{2}\log 2$

$\Rightarrow 2I=I-\frac{\pi }{2}\log 2$

$\Rightarrow I=-\frac{\pi }{2}\log 2$

$\Rightarrow I=-\frac{\pi }{2}\log 2$

$\Rightarrow {\int }_0^{\pi /2}\log \sin xdx=-\frac{\pi }{2}\log 2$.